I worked this problem twice and still did not come up with the correct answer.
The base of a right pyramid is a regular hexagon with sides of length 8m. The altitude is 4m. Find the total surface area of the pyramid.
The right hhexagonal pyramid has 6 identical triangles. The base of the triangle is the same as the side of the base of the pyramid.
I started with the pythagorean theorem.
c^2 = a^2 + b^2
8^2 = 4^2 + b^2
64 = 16 + b^2
-16 -16
48 = b^2
4 square root of 3 = b
Then I used the pythagorean theorem again to find the slant height of the triangles. The legs are 4 square root 3m and 4m.
c^2 = (4 square root of 3)^2 + (4)^2
c^2 = 48 + 16
c^2 = square root of 64
c = 8
So we have a base of 8m and a height of 4m, the area of the triangle is A= 1/2 x bh
1/2 (8)(4) = 16m^2
There are 6 sides, so the lateral area of the pyramid is 6(16) = 96m^2
The base is a regular hexagon with an apothem, a= 4 square root of 3m and perimeter, p=32m. The area of a regular polygon is A = 1/2 AP
The area of the base is 1/2 (4 square root of 3)(32) = 64 square root of 3m^2
The total surface area is the lateral surfaces plus the base.
Answer--->96 + 64 square root of 3m^2
This is how the answer is suppose to be set up, but for some reason I am not coming up with the correct answer. Can someone please tell me where I went wrong in my calculations? Sorry about all the (square root of) stuff, I don't know how to insert the square root symbol
The base of a right pyramid is a regular hexagon with sides of length 8m. The altitude is 4m. Find the total surface area of the pyramid.
The right hhexagonal pyramid has 6 identical triangles. The base of the triangle is the same as the side of the base of the pyramid.
I started with the pythagorean theorem.
c^2 = a^2 + b^2
8^2 = 4^2 + b^2
64 = 16 + b^2
-16 -16
48 = b^2
4 square root of 3 = b
Then I used the pythagorean theorem again to find the slant height of the triangles. The legs are 4 square root 3m and 4m.
c^2 = (4 square root of 3)^2 + (4)^2
c^2 = 48 + 16
c^2 = square root of 64
c = 8
So we have a base of 8m and a height of 4m, the area of the triangle is A= 1/2 x bh
1/2 (8)(4) = 16m^2
There are 6 sides, so the lateral area of the pyramid is 6(16) = 96m^2
The base is a regular hexagon with an apothem, a= 4 square root of 3m and perimeter, p=32m. The area of a regular polygon is A = 1/2 AP
The area of the base is 1/2 (4 square root of 3)(32) = 64 square root of 3m^2
The total surface area is the lateral surfaces plus the base.
Answer--->96 + 64 square root of 3m^2
This is how the answer is suppose to be set up, but for some reason I am not coming up with the correct answer. Can someone please tell me where I went wrong in my calculations? Sorry about all the (square root of) stuff, I don't know how to insert the square root symbol